University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Practice Exercises - Page 474: 39


$\dfrac{\tan^5 x}{5}+c$

Work Step by Step

Consider the integral $\int \tan^4 x \sec^2 x dx$ The given integral can be re-written as: $\int \tan^4 x \sec^2 x dx=\int \tan^4 x (\sec^2 x dx)$ Plug in $a= \tan x \implies da= \sec^2 x dx$ We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$ This implies that $\int a^4 da =\int \dfrac{a^5}{5}+C=\dfrac{\tan^5 x}{5}+c$
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